Question

What is the relation between the wavelength, λ, and the momentum, p, of a particle? Roughly what kinetic energy should an electron have in order for its wavelength be of roughly an "atomic" distance scale? Re-express this as the number of volts needed to accelerate that electron from rest to this energy. Would you call this a high-voltage device?

Answer 1

Answer:

λ = h/p

$E=2.41*10^{-17}Joules$

This energy is equivalent to E=150.44eV

Explanation:

de Broglie Equation:

λ = h/p

h=6.63×10-34 m2 kg / s        Planck's constant

so: p=h/λ

On the other hand, an atomic distance, an angstrom, is equal:$10^{-10}m$, λ should be similar to this distance.

Kinetics Energy of a electron, m=9.11×10−31 kg

$E=\frac{p^{2}}{2m_e}=\frac{h^{2}}{2\lambda^{2}m_e} =\frac{(6.63*10^{-34})^{2}}{2*(10^{-10})^{2}*9.11*10^{-31}}=2.41*10^{-17}Joules$

Number of volts needed to accelerate that electron from rest to this energy:

$E=Vq_{e}$

$V=E/q_{e}=2.41*10^{-17}/(1.602*10^{-19})=150.44Volts$

So, this energy is equivalent to E=150.44eV

This high-voltage device is called: particles accelerator